Question

The simple interest on a sum of money for 2 years at 10% per annum is RS 6000 what will be the compound interest on the same sum at the same rate for the same period .

Don't Spam .​

Answer 1

GIVEN :-

• simple interest , S.I = Rs. 6000.
• Rate , R = 10 % P.a
• Time , T = 2 years.

TO FIND :-

• The compound interest on the same sum at the same rate for the same period.

SOLUTION :-

$\\ \red \bigstar \boxed{ \tt \:S.I = \dfrac{PRT}{100} } \\ \\ \implies \displaystyle \tt \:6000 = \dfrac{P \times 10 \times 2}{100} \\ \\ \implies \displaystyle \tt \:6000 = \dfrac{20P}{100} \\ \\ \implies \displaystyle \tt \:P = \dfrac{6000 \times 100}{20} \\ \\ \implies \displaystyle \tt \:P = \cancel\dfrac{600000}{20} \\ \\ \red \bigstar \: \: \boxed{ \red{ \tt \: P = 30000}} \: \: \red \bigstar$

Now

$\\ \red \bigstar \: \boxed{ \tt \: C.I = P\bigg[ \bigg( 1 + \dfrac{R}{100}\bigg)^{n} - 1 \bigg]} \\ \\ \implies \displaystyle \tt \: C.I = 30000\bigg[ \bigg( 1 + \dfrac{10}{100}\bigg)^{2} - 1 \bigg] \\ \\ \implies \displaystyle \tt \: C.I = 30000\bigg[ \bigg( \dfrac{11}{10}\bigg)^{2} - 1 \bigg] \\ \\ \implies \displaystyle \tt \: C.I = 30000\bigg[ \bigg( \dfrac{121}{100}\bigg) - 1 \bigg] \\ \\ \implies \displaystyle \tt \: C.I = 30000\bigg[ \dfrac{121}{100} - 1 \bigg] \\ \\ \implies \displaystyle \tt \: C.I = 30000\bigg[ \dfrac{121 - 100}{100} \bigg] \\ \\ \implies \displaystyle \tt \: C.I = 30000\bigg[ \dfrac{21}{100} \bigg] \\ \\ \implies \displaystyle \tt \: C.I = 30000 \times \dfrac{21}{100} \\ \\ \implies \displaystyle \tt \: C.I = 300 \times21 \\ \\ \red \bigstar \: \: \boxed{ \red{ \tt \: C.I = 6300}} \: \: \red \bigstar \\ \\ \underline{ \underline{ \therefore \: \tt \: The \: compound \: interest \: is \: Rs. \: 6300}}$

Answer 2

$⤳ \underbrace{ \boxed{ \tt \underline{\underline{ \green{ GIVEN }}}}}$

$\tt⚫S.I. = rs.6000$

$\tt⚫R = 10\%$

$\tt⚫T = 2 \: years$

$⤳ \underbrace{ \boxed{ \tt \underline{\underline{ \pink{ FIND }}}}}$

$\tt⚫C.I. \: on \: the \: same \: sum \: at \: the \: same \\ \tt rate \: for \: the \: same \: period .$

$⤳ \underbrace{ \boxed{ \tt \underline{\underline{ \red{ SOLUTION}}}}}$

$\tt ➔ S.I. = \frac{P\times R \times T}{100}$

$\tt ➔ 6000 = \frac{P\times 10 \times 2}{100}$

$\tt ➔ P = \frac{6000 \times \cancel{10} \cancel0}{1 \cancel0 \times \cancel2} = 60000 \times 5$

$\tt ➔ P = rs.30000$

$\tt☄ Now, C.I.$

$\tt ➨ C.I. = P ({[1+\frac{R}{100}]}^{T}-1)$

$\tt ➨ C.I. = 30000 ({[1+\frac{10}{100}]}^{2}-1)$

$\tt☄ take \: L.C.M$

$\tt ➨ C.I. = 30000 ({[\frac{100 + 10}{100}]}^{2}-1)$

$\tt ➨ C.I. = 30000 ({[\frac{110}{100}]}^{2}-1)$

$\tt ➨ C.I. = 30000 ([1.21]-1)$

$\tt ➨ C.I. = 30000 (0.21)$

$\tt ➨ C.I. = rs.6300$

$\tt \blue{Hence, C.I. on \: the \: same \: sum \: at \: the \: same} \\ \tt\blue {rate \: for \: the \: same \: period \: of \: time} \\ \boxed{ \tt\pink {rs.6300}}$